# Are Velocity and Acceleration Vectors Perpendicular When Speed Is Constant?

• jumboopizza
So, using v^2= u^2 and taking the derivative with respect to time, we get 2\mathbf{v}\cdot\frac{d\mathbf{v}}{dt} = 2u\frac{du}{dt} = 0, which implies that \mathbf{v}\cdot\mathbf{a} = u\frac{du}{dt} = 0. This means that the velocity and acceleration vectors are perpendicular to each other, as desired. In summary, by differentiating the equation v^2= u^2, we can show that the dot product of the velocity and acceleration vectors is equal to the rate of change of speed, which is equal to
jumboopizza

## Homework Statement

the question is simply asking to prove that if speed is constant than velocity and acceleration vectors are perpendicular to each other. it also says as hint to differentiate v dot v= u^2...
V=velocity A=acceleration u=speed t=time

## Homework Equations

i suppose knowing the vector dot product properties would be useful to have around,so here's some:
V dot V= [v]^2(magnitude)
derivative of vector dot products:
d(a dot b)/du= da/du dot b+ a.db/du
acceleration is= the derivative of Velocity with respect to time

## The Attempt at a Solution

ok so I've used the hint and tried to differentiate V dot V= u^2 and i get

dv/dt dot V+V dot dv/dt= du/dt u^2= 0 since speed is constant
2V dot dv/dt=0
now I am not sure if this is representinvg acceleration or not the 2v dv/dt or if its the rate of change in speed or whatever, if its acceleration then wouldn't it be 2A=0, but now it doesn't make sense since acceleration can't be 0 so i know there's a mistake somewhere...

my other idea was to use V dot V=[V]^2 which is one of the dot product properties
and then the magnitude of the velocity would be the speed wouldn't it?

dv/dt [V]^2=0

jumboopizza said:

## Homework Statement

the question is simply asking to prove that if speed is constant than velocity and acceleration vectors are perpendicular to each other. it also says as hint to differentiate v dot v= u^2...
V=velocity A=acceleration u=speed t=time

## Homework Equations

i suppose knowing the vector dot product properties would be useful to have around,so here's some:
V dot V= [v]^2(magnitude)
derivative of vector dot products:
d(a dot b)/du= da/du dot b+ a.db/du
acceleration is= the derivative of Velocity with respect to time

## The Attempt at a Solution

ok so I've used the hint and tried to differentiate V dot V= u^2 and i get

dv/dt dot V+V dot dv/dt= du/dt u^2= 0 since speed is constant
2V dot dv/dt=0
now I am not sure if this is representinvg acceleration or not the 2v dv/dt or if its the rate of change in speed or whatever, if its acceleration then wouldn't it be 2A=0, but now it doesn't make sense since acceleration can't be 0 so i know there's a mistake somewhere...

my other idea was to use V dot V=[V]^2 which is one of the dot product properties
and then the magnitude of the velocity would be the speed wouldn't it?

dv/dt [V]^2=0

acceleration (a vector!) is the time derivative (rate of change) of the velocity vector, so $\mathbf{v} \cdot \frac{d \mathbf{v} }{ dt } = \mathbf{v} \cdot \mathbf{a} = 0$

Under what conditions will two vectors have a dot product of zero?

ok that's what i thought dv/dt= A but its not necessarily 0, if they( vector V and Vecotor A) are perpendicular to eachoter the dot product would be zero...

ok so V dot A=0
so when i first derived V dot V= u^2 and got to dv/dt V+V dv/dt= u^2 du/dt=0

du/dt is the rate of change of speed correct? so if the first derivative of u is =0
would't u'= equal zero?

V dot A= uu' -------------> V dot A= u(0)?

or is it better to make V dot V= [V]^2

sqrt [V]^2= sqrt u^2 ---------> [v]= u

d[v]/dt [v]= du/dt u = 0 --------> d[v]/dt = [a] magnitude of the acceleration

It is better to use $v^2= |\mathbf{v}|^2$ than $u^2$, since $\mathbf{v}\cdot\mathbf{v} = |\mathbf{v}|^2$ follows from the definition of dot product, and by definition, the speed is the magnitude of the velocity.

## 1. What is the equation v dot a= uu' used for in science?

The equation v dot a= uu' is used to calculate the dot product of two vectors, v and a. It is commonly used in physics and engineering to determine the magnitude of a vector in a specific direction.

## 2. How is the dot product represented in this equation?

The dot product is represented by the "dot" symbol between the two vectors, v and a, as in v dot a. It is also sometimes written as v * a.

## 3. What does uu' represent in this equation?

Uu' represents the unit vector of vector u. It is a vector with a magnitude of 1 that points in the same direction as u. In other words, uu' is a way to normalize the vector u.

## 4. How is the dot product calculated in this equation?

The dot product is calculated by multiplying the corresponding components of the two vectors and then summing up the results. For example, if v = [v1, v2, v3] and a = [a1, a2, a3], then v dot a = v1*a1 + v2*a2 + v3*a3.

## 5. What does it mean if the dot product v dot a= uu' equals 0?

If the dot product equals 0, it means that the two vectors, v and a, are perpendicular to each other. This is because the dot product represents the cosine of the angle between the two vectors, and the cosine of 90 degrees (or any multiple of 90 degrees) is 0.

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